When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes 5r/4 Taking the atmospheric pressure to be equal to 10m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

Key ideas needed

1. Hydrostatic Pressure
   Water pressure increases with depth: $P = P_{\text{atm}} + \rho g h$ where $h$ is depth, $\rho$ is water density, and $g$ is acceleration due to gravity.

2. Ideal Gas Law in Constant Temperature (Isothermal)
   For a fixed mass of gas at constant temperature: $P V = \text{constant}$ A bubble is just a tiny gas packet, so if temperature change is ignored, we can write $P_1 V_1 = P_2 V_2$

3. Volume of a Sphere
   $V = \frac{4}{3} \pi r^3$ Volume is proportional to $r^3$.

Logical chain a student would follow

• Step 1: Relate pressures at two positions of the bubble (bottom vs. surface).
• Step 2: Express volumes in terms of the given radii $r$ and $5r/4$.
• Step 3: Use $P_1 V_1 = P_2 V_2$ to link pressure and volume, then cancel common factors.
• Step 4: Insert the known atmospheric pressure (equivalent to 10 m of water) to solve for unknown depth $h$.
• Step 5: Perform quick fraction arithmetic to get an approximate depth.

Each step combines a specific physical principle with a straightforward algebra move, letting you isolate $h$ efficiently.

🫧 Why does the bubble grow?

Imagine you’re holding a balloon deep under water. The deeper you go, the harder the water pushes on it. When you let the balloon float up, that extra push disappears, so the balloon can swell and look bigger.

In this question:

• At the bottom, the water bubble has size r.
• At the top (surface), it grows to 5r/4 (that’s a bit bigger than its old size).
• We know how much the water above pushes (atmospheric pressure is like standing under 10 m of water).

We compare how squeezed the bubble is at the bottom versus the top using a simple rule: "push × space = same number" (that’s the gas law). From that, we can find how deep the bubble started.

It turns out the bubble started roughly 9½ m down.

Step-by-step calculation

1. Pressures
   Bottom of lake: $P_1 = P_{\text{atm}} + \rho g h$
   Surface: $P_2 = P_{\text{atm}}$

2. Volumes
   $V_1 = \frac{4}{3}\pi r^3$
   $V_2 = \frac{4}{3}\pi\left(\frac{5r}{4}\right)^3 = \frac{4}{3}\pi r^3\left(\frac{125}{64}\right)$

3. Isothermal condition
   $P_1 V_1 = P_2 V_2$ $\bigl(P_{\text{atm}} + \rho g h\bigr) V_1 = P_{\text{atm}} \; V_1 \left(\frac{125}{64}\right)$ $P_{\text{atm}} + \rho g h = P_{\text{atm}} \left(\frac{125}{64}\right)$

4. Isolate $h$
   $\rho g h = P_{\text{atm}} \left(\frac{125}{64} - 1\right) = P_{\text{atm}} \left(\frac{61}{64}\right)$

5. Insert atmospheric pressure in water-column form
   Given $P_{\text{atm}} = \rho g (10\,\text{m})$
   $\rho g h = \rho g (10) \left(\frac{61}{64}\right)$
   $h = 10 \times \frac{61}{64} \approx 9.53\,\text{m}$

Final Answer

The depth of the lake is approximately $9.5\,\text{m}$.